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STRUCTURE OF OXYGEN ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories which could not lead to the correct nuclear structure. For example the paper “The Cluster Structure of Oxygen Isotopes “ cannot lead to the nuclear structure. . Under this physics crisis and using the charged Up and Down quarks, discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for discovering the nuclear force and structure by applying the laws of electromagnetism.(See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). Naturally occurring oxygen is composed of three stable isotopes, O-16, O-17, and O-18, with O-16 being the most abundant (99.762% natural abundance); thus oxygen (O) has a standard atomic weight of 15.9994(3). Known oxygen isotopes range in mass number from 12 to 24. Most O-16 is synthesized at the end of the helium fusion process in stars; the triple-alpha reaction creates C-12, which captures an additional He-4 to make O-16. The neon burning process creates additional O-16. Both O-17 and O-18 are secondary isotopes, meaning that their nucleosynthesis requires seed nuclei. O-17 is primarily made by the burning of hydrogen into helium during the CNO cycle, making it a common isotope in the hydrogen burning zones of stars. Most O-18 is produced when N-14 (made abundant from CNO burning) captures a He-4 nucleus, making O-18 common in the helium-rich zones of stars. In order to reveal the structure of oxygen isotopes I start with the diagram of my paper STRUCTURE OF O-16 AND O-15 In the following diagram of the stable O-16 you see that the 4 nucleons of the first horizontal plane have positive spins (+HP1), while the 4 nucleons of the fourth horizontal plane have 4 nucleons of negative spins (-HP4). ' DIAGRAM OF O-16' ' ' ' p8(-1/2).n8(-1/2)' n7(-1/2).p7(-1/2) -HP4 ' n6(+1/2).p6(+1/2)' p5(+1/2).n5(+1/2) +HP3 ' p4(-1/2).n4(-1/2)' ' n3 (-1/2).p3 (- 1/2). –HP2 ' ' n2(+1/2).p2(+1/2). ' ' p1(+1/2). n1(+1/2) +HP1' For understanding better the structure of oxygen-16 you can see the third following figure STRUTURE OF O-12 WITH S= 0 ''' Using the diagram of O-16 with S=0 we see that the absence of 4 neutrons of opposite spins (one neutron per horizontal plane) give the structure of O-12 with the same S = 0. It is similar to the structure of O-16 but here the small number of pn bonds of short range cannot overcome the pp repulsions of long range. For example the absent 4 neutrons as n1, n3, n6 and n8 of opposite spins reduce the np bonds and the pp repulsions of long range lead to the decay. '''STRUCTURE OF O-13 WITH S =-3/2 Here the absent 3 neutrons which form the O-13 give a negative spin S =-3/2. It means that the nucleons p1n1p2n2 of the square of the first horizontal plane of positive spins (+HP1) are moved to make an alpha particle near the p3n4n5p6. Thus in the present of all 8 neutrons the total spin is given by The second plane of 6 nucleons gives S=-3 The third plane of 6 nucleons gives S =+3 The fourth plane of 4 nucleons gives S =-2 Thus in the absence of 3 neutrons (one in each plane) we get S = -3/2 of the 3 nucleons of the fourth plane, while the 10 nucleons of the second and the third plane give S=0. STRUCTURE OF O-14 AND O-15 In the absence of the n1(+1/2) and the n8(-1/2) we get the O-14 with S=0. It has a similar structure of O-16 but here the absence of 2 neutrons reduces the pn bonds of short range. Then such a structure cannot overcome the pp repulsions of long range. Also in the absence of n1(+1/2) which gives the O-15 of S=-1/2 the reduced pn bonds of short range cannot overcome the pp repulsions of long range. (See my STRUCTURE OF O-16 AND O-15 ). STRUCTURE OF O-17 WITH S = +5/2 Because of this great positive spin of the O-17 here the 12 nucleons have the structure of carbon in which the nucleons p7n7p8n8 make a square of positive spins existing over the structure of carbon as shown in the following diagram ' DIAGRAM OF O-17' ' n7(+1/2)..........p8(+1/2).........n' ' p7(+1/2) .........n8(+1/2 +HP3' ' p2(-1/2)..........n4 (-1/2)……….p6(-1/2) ' ' n2 (-1/2)..........p4 (- 1/2)……….n6(-1/2) -HP2 ' ' n1(+1/2)..........p3(+1/2)……….n5(+1/2) ' ' p1(+1/2)..........n3(+1/2)……….p5(+1/2) +Hp1 ' In this diagram one sees that the nucleons from p1 to n6 give S=0 while the nucleons of the third horizontal plane give S= +5/2 which is the total spin of O-17. Here you also see that the extra neutron n makes not a single bond but two bonds per neutron. Especially it makes a weak horizontal bond with p8 and a vertical strong bond with p6. Under this condition the O-17 is a stable nuclide. STRUCTURE OF O-18 WITH S =0 Here using the diagram of O-16 we suggest that it has the two extra neutrons like n9 and n10 of opposite spins which make two horizontal single bonds with p3(-1/2) and p5(+1/2). Therefore the p3 makes 5 bonds per proton like the p3n9, the p3n1 the p3n5, the p3n3 and the p3n4. Similarly the p5 makes also 5 bonds per proton. Thus in the presence of 5 bonds per proton the single bonds p3n9 and p5n10 have binding energies able to overcome the nn repulsions of short range. Under this condition the O-18 is a stable nuclide. STRUCTURE OF O-19 WITH S= +5/2 Here using the diagram of O-17 we see that the O-19 has not only the extra n of strong bonds but also two extra neutrons of opposite spins which make two single bonds with p3 and p4. They are not shown in the diagram of O-17 because they exist behind the p3 and in front of p4. Note that here the three extra neutrons contribute more to the nn repulsions than the two extra neutrons of O-18 and lead to the beta decay. STRUCTURE OF O-20 , O-22 AND O-24 WITH S=0 In these cases using the diagram of O-16 we see that the O-20 with S =0 consists of 4 extra neutron of opposite spins with single bonds (one extra neutron in each horizontal plane) unable to overcome the nn repulsions of short range. Moreover the O-24 consists of 6 extra neutrons with opposite spins. Similarly the O-24 consists of 8 extra neutrons of opposite spins. STRUCTURE OF O-21 and O-23 WITH S = +1/2 Here the O-21 with S = +1/2 has 5 extra neutrons of single bonds giving a total spin S =+1/2. Moreover the O-23 with S =+1/2 has 7 extra neutrons of single bonds giving the same total spin S =+1/2. Category:Fundamental physics concepts